I will write $R$ for $R_{q}$, because $q$ is constant. In the following, I am
going to assume that $R$ is an *arbitrary* right inverse of $M_{q}$, rather
than the specific right inverse $M_{q}^{T}\left(  M_{q}M_{q}^{T}\right)
^{-1}$ which you have suggested. This makes the statement a tad more general
and rids us of a red herring.

I shall prove that for any
$i\in\left\{  1,2,...,n\right\}  $, $j\in\left\{  1,2,...,n\right\}  $ and
$k\in\left\{  1,2,...,n-1\right\}  $, the value $C_{qi}R_{jk}-C_{qj}R_{ik}$ is
a homogeneous polynomial of degree $n-2$ in the entries of the matrix $M_{q}$
(independently, and independent, of the choice of right inverse $R$).

For every $\ell\in\left\{  1,2,...,n\right\}  $, we abbreviate $C_{q\ell}$ by
$v_{q}$. Thus,

$v_{\ell}=C_{q\ell}=\left(  -1\right)  ^{q+\ell}\det\left(
\underbrace{M\text{ without row }q}_{=M_{q}}\text{ and column }\ell\right)  $

**(1)** $=\left(  -1\right)  ^{q+\ell}\det\left(  M_{q}\text{ without column
}\ell\right)  $.

From this it easy to obtain

**(2)** $\sum_{\ell\in\left\{  1,2,...,n\right\}  }\left(  M_{q}\right)
_{j\ell}v_{\ell}=0$ for every $j \in \left\{1,2,...,n-1\right\}$.

[*Proof of **(2)**:* Let $j \in \left\{1,2,...,n-1\right\}$. Let $G$ be the result of inserting a copy of row $j$ of the matrix $M_{q}$ between the rows $q-1$ and $q$ of this matrix. Then, $G$ is an $n\times n$-matrix with two equal rows, and hence has determinant $\det G=0$. But Laplace expansion of $\det G$ along the row we have inserted yields

$\det G=\sum_{\ell\in\left\{  1,2,...,n\right\}  }\left(  M_{q}\right)
_{j\ell}\left(  -1\right)  ^{q+\ell}\det\left(  M_{q}\text{ without column
}\ell\right)  $,

since the entries of this row are $\left(  M_{q}\right)  _{j\ell}$ and the
cofactors corresponding to this row are $\left(  -1\right)  ^{q+\ell}
\det\left(  M_{q}\text{ without column }\ell\right)  $. Now, **(1)** yields

$\sum_{\ell\in\left\{  1,2,...,n\right\}  }\left(  M_{q}\right)  _{j\ell
}v_{\ell}$

$=\sum_{\ell\in\left\{  1,2,...,n\right\}  }\left(  M_{q}\right)  _{j\ell
}\left(  -1\right)  ^{q+\ell}\det\left(  M_{q}\text{ without column }
\ell\right)  $

$=\det G=0$,

so that **(2)** is proven.]

Since $M$ is generic, its cofactors $v_{1}$, $v_{2}$, $...$, $v_{n}$ are nonzero.

Fix $i\in\left\{  1,2,...,n\right\}  $. Let $N_{i}$ denote the matrix $M_{q}$
without column $i$. Thus, $\det\left(  N_{i}\right)  =\det\left(  M_{q}\text{
without column }i\right)  =\left(  -1\right)  ^{q+i}v_{i}$ (because **(1)**
yields $v_{i}=\left(  -1\right)  ^{q+i}\det\left(  M_{q}\text{ without column
}i\right)  $).

Let $S_{i}$ denote the matrix $\left(  R_{jk}-\dfrac{v_{j}}{v_{i}}
R_{ik}\right)  _{j\in\left\{  1,2,...,n\right\}  ;\ k\in\left\{
1,2,...,n-1\right\}  }$. The $i$-th row of this matrix $S_{i}$ is zero; let
$S_{i}^{\prime}$ denote the matrix $S_{i}$ without row $i$.

Now, we claim that $N_{i}S_{i}^{\prime}=I_{n-1}$ (the $\left(  n-1\right)
\times\left(  n-1\right)  $ identity matrix). Indeed, for every $\left(
j,k\right)  \in\left\{  1,2,...,n-1\right\}  ^{2}$, the $\left(  j,k\right)
$-th entry of $N_{i}S_{i}^{\prime}$ is

$\left(  N_{i}S_{i}^{\prime}\right)  _{jk}=\sum_{u\in\left\{
1,2,...,n-1\right\}  }\left(  N_{i}\right)  _{ju}\left(  S_{i}^{\prime
}\right)  _{uk}$

$=\sum_{\ell\in\left\{  1,2,...,n\right\}  \setminus\left\{  i\right\}
}\left(  M_{q}\right)  _{j\ell}\underbrace{\left(  S_{i}\right)  _{\ell k}
}_{=R_{\ell k}-\dfrac{v_{\ell}}{v_{i}}R_{ik}}$

(since $N_{i}$ is the matrix $M_{q}$ without column $i$, while $S_{i}^{\prime
}$ is the matrix $S_{i}$ without row $i$)

$=\sum_{\ell\in\left\{  1,2,...,n\right\}  \setminus\left\{  i\right\}
}\left(  M_{q}\right)  _{j\ell}\left(  R_{\ell k}-\dfrac{v_{\ell}}{v_{i}
}R_{ik}\right)  $

$=\sum_{\ell\in\left\{  1,2,...,n\right\}  }\left(  M_{q}\right)  _{j\ell
}\left(  R_{\ell k}-\dfrac{v_{\ell}}{v_{i}}R_{ik}\right)  $

(here, we added an $\ell=i$ addend to the sum; this did not change the sum
because this addend is $0$)

$=\underbrace{\sum_{\ell\in\left\{  1,2,...,n\right\}  }\left(  M_{q}\right)
_{j\ell}R_{\ell k}}_{\substack{=\left(  M_{q}R\right)  _{jk}=\delta
_{jk}\\\text{(since }R\text{ is a}\\\text{right inverse of }M_{q}\text{)}
}}-\underbrace{\sum_{\ell\in\left\{  1,2,...,n\right\}  }\left(  M_{q}\right)
_{j\ell}\dfrac{v_{\ell}}{v_{i}}R_{ik}}_{=\dfrac{1}{v_{i}}R_{ik}\sum_{\ell
\in\left\{  1,2,...,n\right\}  }\left(  M_{q}\right)  _{j\ell}v_{\ell}}$

$=\delta_{jk}-\dfrac{1}{v_{i}}R_{ik}\underbrace{\sum_{\ell\in\left\{
1,2,...,n\right\}  }\left(  M_{q}\right)  _{j\ell}v_{\ell}}
_{\substack{=0\\\text{(by \textbf{(2)})}}}=\delta_{jk}$.

This shows that $N_{i}S_{i}^{\prime}=I_{n-1}$, and thus $S_{i}^{\prime}
=N_{i}^{-1}$ (since $N_{i}$ and $S_{i}^{\prime}$ are $\left(  n-1\right)
\times\left(  n-1\right)  $-matrices).

Now, we recall Cramer's rule for the inverse of a matrix. It essentially says
that the inverse of a square matrix is obtained by dividing its adjoint by its
determinant. In other words, if $X$ is an invertible $p\times p$-matrix for
some $p\in\mathbb{N}$, and $j$ and $k$ are elements of $\left\{
1,2,...,p\right\}  $, then

**(3)** $\left(  X^{-1}\right)  _{jk}=\dfrac{1}{\det X}\left(  -1\right)
^{j+k}\det\left(  X\text{ without row }k\text{ and column }j\right)  $.

Now, recall that $S_{i}^{\prime}=N_{i}^{-1}$. Hence,

$\left(  S_{i}^{\prime}\right)  _{jk}=\left(  N_{i}^{-1}\right)  _{jk}
=\dfrac{1}{\det\left(  N_{i}\right)  }\left(  -1\right)  ^{j+k}\det\left(
N_{i}\text{ without row }k\text{ and column }j\right)  $ (by **(3)**)

$=\dfrac{1}{\left(  -1\right)  ^{q+i}v_{i}}\left(  -1\right)  ^{j+k}
\det\left(  N_{i}\text{ without row }k\text{ and column }j\right)  $ (since
$\det\left(  N_{i}\right)  =\left(  -1\right)  ^{q+i}v_{i}$)

$=\dfrac{1}{v_{i}}\left(  -1\right)  ^{i+j+q+k}\det\left(  N_{i}\text{ without
row }k\text{ and column }j\right)  $

for all $\left(  j,k\right)  \in\left\{  1,2,...,n-1\right\}  ^{2}$. In other words,

**(4)** $v_{i}\left(  S_{i}^{\prime}\right)  _{jk}=\left(  -1\right)
^{i+j+q+k}\det\left(  N_{i}\text{ without row }k\text{ and column }j\right)  $

for all $\left(  j,k\right)  \in\left\{  1,2,...,n-1\right\}  ^{2}$.

Let $\mathbf{B}$ be the (unique) increasing bijection from $\left\{
1,2,...,n\right\}  \setminus\left\{  i\right\}  $ to $\left\{
1,2,...,n-1\right\}  $. Then, for all $j\in\left\{  1,2,...,n\right\}
\setminus\left\{  i\right\}  $ and $k\in\left\{  1,2,...,n-1\right\}  $, we have

**(5)** $v_{i}\left(  S_{i}\right)  _{jk}=\left(  -1\right)  ^{i+\mathbf{B}
\left(  j\right)  +q+k}\det\left(  M_{q}\text{ without row }k\text{ and
columns }i\text{ and }j\right)  $.

(Indeed, this follows from **(4)**, applied to $\mathbf{B}\left(  j\right)  $
instead of $j$, because $\left(  S_{i}^{\prime}\right)  _{\mathbf{B}\left(
j\right)  ,k}=\left(  S_{i}\right)  _{jk}$ (since $S_{i}^{\prime}$ is the
matrix $S_{i}$ without row $i$) and because column $j$ of $M_{q}$ is column
$\mathbf{B}\left(  j\right)  $ of $N_{i}$ (since $N_{i}$ is the matrix $M_{q}$
without column $i$).)

Now, fix $j\in\left\{  1,2,...,n\right\}  $ and $k\in\left\{
1,2,...,n-1\right\}  $. We need to show that $C_{qi}R_{jk}-C_{qj}R_{ik}$ is a
homogeneous polynomial of degree $n-2$ in the entries of the matrix $M_{q}$.
We WLOG assume that $j\neq i$ (since otherwise, $C_{qi}R_{jk}-C_{qj}R_{ik}
=0$). Thus, $j\in\left\{  1,2,...,n\right\}  \setminus\left\{  i\right\}  $.
Since $C_{qi}=v_{i}$ and $C_{qj}=v_{j}$ (by the definitions of $v_{i}$ and
$v_{j}$), we have

$\underbrace{C_{qi}}_{=v_{i}}R_{jk}-\underbrace{C_{qj}}_{=v_{j}}R_{ik}
=v_{i}R_{jk}-v_{j}R_{ik}=v_{i}\underbrace{\left(  R_{jk}-\dfrac{v_{j}}{v_{i}
}R_{ik}\right)  }_{\substack{=\left(  S_{i}\right)  _{jk}\\\text{(by the
definition of }S_{i}\text{)}}}$

**(6)** $=v_{i}\left(  S_{i}\right)  _{jk}=\left(  -1\right)  ^{i+\mathbf{B}\left(
j\right)  +q+k}\det\left(  M_{q}\text{ without row }k\text{ and columns
}i\text{ and }j\right)  $

(by **(5)**),

which is obviously a homogeneous polynomial of degree $n-2$ in the entries of
the matrix $M_{q}$ (and independent of the choice of $R$), qed.

Thanks for a very nice question, and sorry for this mess of an answer...

**PS.** I believe the genericity of $M$ is not required for **(6)** to hold. Does anyone see a nice proof of this? I don't.